Chemistry, 12.12.2019 05:31 williamslyric

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm3. if the sphere were broken down into eight spheres each having a volume of 1.25 cm3, and the reaction is run a second time, which of the following accurately characterizes the second run? choose all that apply a. the second run will be faster. b. the second run will be slower. c. the second run will have the same rate as the first. d. the second run has twice the surface area. e.the second run has eight times the surface area. f. the second run has 10 times the surface area

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Answer from: kyle65

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reactionThe second run will be slower - false, the increased surface area of catalyst will increase the rate of reactionThe second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reactionThe second run has twice the surface area - yes, 44 sqcm to 22 sqcmThe second run has eight times the surface area - no, 44 sqcm to 22 sqcmThe second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

Answer from: Quest

answer is: naturally occurring solid, inorganic compound or element is mineral.

minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

for example magnetite is a mineral of iron oxide (fe₂o₃).

calcite is a carbonate mineral and the most stable polymorph of calcium carbonate (caco₃).

dolomite is an carbonate mineral composed of calcium magnesium carbonate (camg(co₃)₂).

Answer from: Quest

Must be negative.substance increases

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